Apply linear & quadratic equations

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} - 4$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{−4}$. The coordinates of this point are $\left(0; -4\right)$.

The $x$-intercepts are the roots of the equation $x^{2} - 4 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{0}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{4}$. The two numbers are $\text{2}$ and $-\text{2}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = 2$ and $x = -2$. The coordinates of these roots are $\left(2; 0 \right)$ and $\left(-2; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $\text{2}$. Its coordinates are: $\left(0; 2\right)$.

We will work out the root of the equation $0 = - x + 2$.

The root is: $x = 2$. It's coordinates are: $\left( 2; 0 \right)$.

We will now plot the equation $y = - x + 2$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(2; 0)$ and $(-3; 5)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} - 4$ and $y = - x + 2$. The solutions are the points where their graphs intersect.

The points of intersection are $(2; 0)$ and $(-3; 5)$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} + 2 x - 3$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{−3}$. The coordinates of this point are $\left(0; -3\right)$.

The $x$-intercepts are the roots of the equation $x^{2} + 2 x - 3 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{2}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{3}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{3}$. The two numbers are $\text{3}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = 1$ and $x = -3$. The coordinates of these roots are $\left(1; 0 \right)$ and $\left(-3; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $\text{1}$. Its coordinates are: $\left(0; 1\right)$.

We will work out the root of the equation $0 = - x + 1$.

The root is: $x = 1$. It's coordinates are: $\left( 1; 0 \right)$.

We will now plot the equation $y = - x + 1$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(1; 0)$ and $(-4; 5)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 2 x - 3$ and $y = - x + 1$. The solutions are the points where their graphs intersect.

The points of intersection are $(1; 0)$ and $(-4; 5)$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} - 7 x + 6$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{6}$. The coordinates of this point are $\left(0; 6\right)$.

The $x$-intercepts are the roots of the equation $x^{2} - 7 x + 6 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{7}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{6}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{6}$. The two numbers are $-\text{1}$ and $-\text{6}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = 6$ and $x = 1$. The coordinates of these roots are $\left(6; 0 \right)$ and $\left(1; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $-\text{8}$. Its coordinates are: $\left(0; -8\right)$.

We will work out the root of the equation $0 = 2 x - 8$.

The root is: $x = 4$. It's coordinates are: $\left( 4; 0 \right)$.

We will now plot the equation $y = 2 x - 8$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(7; 6)$ and $(2; -4)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} - 7 x + 6$ and $y = 2 x - 8$. The solutions are the points where their graphs intersect.

The points of intersection are $(7; 6)$ and $(2; -4)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $y$.

Now we need to substitute the entire right side of the equation, $(2 x - 16)$, into the quadratic equation in place of $y$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $x$-values.

Cool! However, there is still work to do. We must now find the values of $y$ for these equations. We must substitute in the values $x = 1$ and $x = -4$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = 1$ then $y = -14$ and if $x = -4$ then $y = -24$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(1 ; -14\right)$ and $\left(-4 ; -24\right)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $x$.

Now we need to substitute the entire right side of the equation, $(3 y + 6)$, into the quadratic equation in place of $x$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$-values.

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = -2$ and $y = 2$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = 0$ then $y = -2$ and if $x = 12$ then $y = 2$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(0 ; -2\right)$ and $\left(12 ; 2\right)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $x$.

Now we need to substitute the entire right side of the equation, $(- 6 y - 3)$, into the quadratic equation in place of $x$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$-values.

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = 4$ and $y = 0$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = -27$ then $y = 4$ and if $x = -3$ then $y = 0$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(-27 ; 4\right)$ and $\left(-3 ; 0\right)$.

You need to start by substituting $- 2 x + 5$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{5}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{4}$. The two numbers are $-\text{1}$ and $-\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{4}$ and $\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = 4$:

For $x = 1$:

The solutions are $x = 4$ together with $y = -3$ , and $x = 1$ with $y = 3$. In coordinates pairs, the solutions are $\left(4; -3\right)$ and $\left(1; 3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; -3\right)$ and $\left(1; 3\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; -3\right)$ and $\left(1; 3\right)$.

You need to start by substituting $- x - 1$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{5}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{4}$. The two numbers are $\text{1}$ and $\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $-\text{1}$ and $-\text{4}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = -1$:

For $x = -4$:

The solutions are $x = -1$ together with $y = 0$ , and $x = -4$ with $y = 3$. In coordinates pairs, the solutions are $\left(-1; 0\right)$ and $\left(-4; 3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; 0\right)$ and $\left(-4; 3\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-1; 0\right)$ and $\left(-4; 3\right)$.

You need to start by substituting $- x - 1$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{8}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{8}$. The two numbers are $\text{2}$ and $\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $-\text{2}$ and $-\text{4}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = -2$:

For $x = -4$:

The solutions are $x = -2$ together with $y = 1$ , and $x = -4$ with $y = 3$. In coordinates pairs, the solutions are $\left(-2; 1\right)$ and $\left(-4; 3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-2; 1\right)$ and $\left(-4; 3\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-2; 1\right)$ and $\left(-4; 3\right)$.

You need to start by substituting $- x + 3$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{1}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{12}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $-\text{4}$. The two numbers are $\text{3}$ and $-\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\dfrac{4}{3}$ and $-\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = \dfrac{4}{3}$:

For $x = -1$:

The solutions are $x = \dfrac{4}{3}$ together with $y = \dfrac{5}{3}$ , and $x = -1$ with $y = 4$. In coordinate pairs, the solutions are $\left(\dfrac{4}{3}; \dfrac{5}{3}\right)$ and $\left(-1; 4\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{4}{3}; \dfrac{5}{3}\right)$ and $\left(-1; 4\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{4}{3}; \dfrac{5}{3}\right)$ and $\left(-1; 4\right)$.

You need to start by substituting $- x - 3$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{11}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{24}$. This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{12}$. The two numbers are $\text{3}$ and $\text{8}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $-\dfrac{3}{2}$ and $-\text{4}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = -\dfrac{3}{2}$:

For $x = -4$:

The solutions are $x = -\dfrac{3}{2}$ together with $y = -\dfrac{3}{2}$ , and $x = -4$ with $y = 1$. In coordinate pairs, the solutions are $\left(-\dfrac{3}{2}; -\dfrac{3}{2}\right)$ and $\left(-4; 1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-\dfrac{3}{2}; -\dfrac{3}{2}\right)$ and $\left(-4; 1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-\dfrac{3}{2}; -\dfrac{3}{2}\right)$ and $\left(-4; 1\right)$.

You need to start by substituting $2 x + 3$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{5}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{6}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $-\text{2}$. The two numbers are $\text{6}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\dfrac{1}{3}$ and $-\text{2}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = \dfrac{1}{3}$:

For $x = -2$:

The solutions are $x = \dfrac{1}{3}$ together with $y = \dfrac{11}{3}$ , and $x = -2$ with $y = -1$. In coordinate pairs, the solutions are $\left(\dfrac{1}{3}; \dfrac{11}{3}\right)$ and $\left(-2; -1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{1}{3}; \dfrac{11}{3}\right)$ and $\left(-2; -1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{1}{3}; \dfrac{11}{3}\right)$ and $\left(-2; -1\right)$.

You need to start by substituting $y - 2$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{1}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{2}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{2}$. The two numbers are $\text{1}$ and $-\text{2}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 2 and -1.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 2$:

For $y = -1$:

The solutions are $x = 0$ together with $y = 2$ , and $x = -3$ with $y = -1$. In coordinates pairs, the solutions are $\left(0; 2\right)$ and $\left(-3; -1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(0; 2\right)$ and $\left(-3; -1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(0; 2\right)$ and $\left(-3; -1\right)$.

You need to start by substituting $y - 1$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{5}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{4}$. The two numbers are $-\text{1}$ and $-\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 4 and 1.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 4$:

For $y = 1$:

The solutions are $x = 3$ together with $y = 4$ , and $x = 0$ with $y = 1$. In coordinates pairs, the solutions are $\left(3; 4\right)$ and $\left(0; 1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(3; 4\right)$ and $\left(0; 1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(3; 4\right)$ and $\left(0; 1\right)$.

You need to start by substituting $y + 5$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{4}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{3}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{3}$. The two numbers are $\text{1}$ and $\text{3}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are -1 and -3.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = -1$:

For $y = -3$:

The solutions are $x = 4$ together with $y = -1$ , and $x = 2$ with $y = -3$. In coordinates pairs, the solutions are $\left(4; -1\right)$ and $\left(2; -3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; -1\right)$ and $\left(2; -3\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; -1\right)$ and $\left(2; -3\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $- \frac{x}{2} + 2$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{4}$. So multiply both sides of the equation by $\text{4}$ in order to cancel the denominators. Then arrange the equation in standard form.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -3$ and $x = 4$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-3 ; \frac{7}{2}\right)$ and $\left(4 ; 0\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-3 ; \frac{7}{2}\right)$ and $\left(4 ; 0\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $- \frac{x}{3} + \frac{4}{3}$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{3}$. So multiply both sides of the equation by $\text{3}$ in order to cancel the denominators. Then arrange the equation in standard form.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -2$ and $x = 4$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-2 ; 2\right)$ and $\left(4 ; 0\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-2 ; 2\right)$ and $\left(4 ; 0\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $\frac{20 x}{3} + 16$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{3}$. So multiply both sides of the equation by $\text{3}$ in order to cancel the denominators. Then arrange the equation in standard form.

In this case, there is a common factor of $\text{−4}$ for all of the terms. Divide out this factor before factorising.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -4$ and $x = 0$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-4 ; - \frac{32}{3}\right)$ and $\left(0 ; 16\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-4 ; - \frac{32}{3}\right)$ and $\left(0 ; 16\right)$.

You need to substitute $- 3 x + 9$ in place of $y$ in the quadratic equation $y = x^{2} - x + 6$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $- 3 x + 9$ from the linear equation in place of $y$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $- 3 x + 9=x^{2} - x + 6$ which is choice A.

You need to substitute $y + 3$ in place of $x$ in the quadratic equation $y = x^{2} + 2 x - 5$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $y + 3$ from the linear equation in place of $x$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $y$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $y=(y + 3)^2+2(y + 3)-5$ which is choice B.

You need to substitute $- 2 x + 2$ in place of $y$ in the quadratic equation $y = x^{2} - 4 x - 1$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $- 2 x + 2$ from the linear equation in place of $y$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $- 2 x + 2=x^{2} - 4 x - 1$ which is choice D.

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $xy$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $xy$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by $\text{2}$ so that the $x$-coefficients will be ready to cancel. Then we can eliminate them by adding the equations.

First modify things to set up the elimination:

Now do the elimination and complete the solution to find $y$.

Great - we have the first value, $y = \frac{1}{2}$.

Now the last step: find the value of $x$. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

The answer is the pair of numbers $x = \frac{1}{4}$ and $y = \frac{1}{2}$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $x = \frac{1}{4}$ and $y = \frac{1}{2}$.

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $ab$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $ab$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations invite elimination, because the $b$-terms have equal and opposite coefficients. We can eliminate the $b$-terms by adding the equations.

Eliminate the $b$ and solve for $a$:

Great - we have the first value, $a = 9$.

Now the last step: find the value of $b$. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

The answer is the pair of numbers $a = 9$ and $b = -4$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $a = 9$ and $b = -4$.

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $xy$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $xy$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by $\text{2}$ so that the $y$-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

Now do the elimination and complete the solution to find $x$.

Great - we have the first value, $x = 7$.

Now the last step: find the value of $y$. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

The answer is the pair of numbers $x = 7$ and $y = 6$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $x = 7$ and $y = 6$.

You need to start by substituting $- x - 1$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = 3 x^{2} + 5 x + 2$ and the linear equation is $y = - x - 1$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{9}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{3}$. If we take $\text{3}$ which is half of the value of $\text{6}$ (the coefficient of the middle term) and square it, we get the product $\text{9}$. In this case we will use only one number to factorise, that is, $\text{3}$. We will re-write the middle term, forming two terms, using this value.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $-\text{1}$. This root is the $x$-value of our solution. This $x$-value will have one corresponding $y$ value.

Now that we have the value of $x$, we will substitute it into the linear equation to calculate the corresponding value of $y$.

The solution is $x = -1$ together with $y = 0$. Using coordinate pairs, the solution is $\left(-1; 0\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; 0\right)$. This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(-1; 0\right)$.

You need to start by substituting $- 3 x - 3$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = 2 x^{2} + x - 1$ and the linear equation is $y = - 3 x - 3$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $\text{4}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{4}$. This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{2}$. If we take $\text{2}$ which is half of the value of $\text{4}$ (the coefficient of the middle term) and square it, we get the product $\text{4}$. In this case we will use only one number to factorise, that is, $\text{2}$. We will re-write the middle term, forming two terms, using this value.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $-\text{1}$. This root is the $x$-value of our solution. This $x$-value will have one corresponding $y$ value.

Now that we have the value of $x$, we will substitute it into the linear equation to calculate the corresponding value of $y$.

The solution is $x = -1$ together with $y = 0$. Using coordinate pairs, the solution is $\left(-1; 0\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; 0\right)$. This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(-1; 0\right)$.

You need to start by substituting $- 2 x - 3$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = x^{2} + 4 x + 6$ and the linear equation is $y = - 2 x - 3$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{9}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{9}$. If we take $\text{3}$ which is half of the value of $\text{6}$ (the coefficient of the middle term) and square it, we get the product $\text{9}$. In this case we will use only one number to factorise, that is, $\text{3}$. We will re-write the middle term, forming two terms, using this value.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $-\text{3}$. This root is the $x$-value of our solution. This $x$-value will have one corresponding $y$ value.

Now that we have the value of $x$, we will substitute it into the linear equation to calculate the corresponding value of $y$.

The solution is $x = -3$ together with $y = 3$. Using coordinate pairs, the solution is $\left(-3; 3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-3; 3\right)$. This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(-3; 3\right)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $4 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = 4 y$.

For $y = 2$:

And for $y = -2$:

So the $x$-values are $x = 8 \text{ or } x = -8$.

Therefore the correct coordinate pairs are $(8; 2)$ and $(-8; -2)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $- 2 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = - 2 y$.

For $y = 2$:

And for $y = -2$:

So the $x$-values are $x = -4 \text{ or } x = 4$.

Therefore the correct coordinate pairs are $(-4; 2)$ and $(4; -2)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $2 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = 2 y$.

For $y = 3$:

And for $y = -3$:

So the $x$-values are $x = 6 \text{ or } x = -6$.

Therefore the correct coordinate pairs are $(6; 3)$ and $(-6; -3)$.

You need to start by substituting $- y + 3$ in place of $x$ in the hyperbolic equation and then solve for $y$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $x$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are $\text{2}$ and $\text{8}$.

By using the values of $y$, we will calculate the corresponding values of $x$. We will do so by substituting each $y$-value into the linear equation.

For $y = 2$:

For $y = 8$:

The solutions are $x = 1$ together with $y = 2$ , and $x = -5$ with $y = 8$. Using coordinate pairs, the solutions are $\left(1; 2\right)$ and $\left(-5; 8\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(1; 2\right)$ and $\left(-5; 8\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(1; 2\right)$ and $\left(-5; 8\right)$.

You need to start by substituting $x + 4$ in place of $y$ in the hyperbolic equation and then solve for $x$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $-\text{2}$ and $\text{3}$.

By using the values of $x$, we will calculate the corresponding values of $y$. We will do so by substituting each $x$-value into the linear equation.

For $x = -2$:

For $x = 3$:

The solutions are $x = -2$ together with $y = 2$ , and $x = 3$ with $y = 7$. Using coordinate pairs, the solutions are $\left(-2; 2\right)$ and $\left(3; 7\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-2; 2\right)$ and $\left(3; 7\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-2; 2\right)$ and $\left(3; 7\right)$.

You need to start by substituting $- y + 1$ in place of $x$ in the hyperbolic equation and then solve for $y$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $x$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are $-\text{3}$ and $\text{1}$.

By using the values of $y$, we will calculate the corresponding values of $x$. We will do so by substituting each $y$-value into the linear equation.

For $y = -3$:

For $y = 1$:

The solutions are $x = 4$ together with $y = -3$ , and $x = 0$ with $y = 1$. Using coordinate pairs, the solutions are $\left(4; -3\right)$ and $\left(0; 1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; -3\right)$ and $\left(0; 1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; -3\right)$ and $\left(0; 1\right)$.

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

The correct equation for the first fact is $n_{1} \times n_{2} = 90$.

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like $\text{10}$ and $\text{11}$, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is $\text{1}$ more than the small number." Another is "the small number is $\text{1}$ less than the larger number." Let's use the second option.

This makes sense: consecutive numbers are separated by $\text{1}$. The equation shows that with the $-\text{1}$.

The correct answers are:

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

The correct equation for the first fact is $n_{1} \times n_{2} = 56$.

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like $\text{10}$ and $\text{11}$, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is $\text{1}$ more than the small number." Another is "the small number is $\text{1}$ less than the larger number." Let's use the first option.

This makes sense: consecutive numbers are separated by $\text{1}$. The equation shows that with the $\text{+1}$.

The correct answers are:

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

The correct equation for the first fact is $n_{1} \times n_{2} = 132$.

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like $\text{10}$ and $\text{11}$, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is $\text{1}$ more than the small number." Another is "the small number is $\text{1}$ less than the larger number." Let's use the first option.

This makes sense: consecutive numbers are separated by $\text{1}$. The equation shows that with the $\text{+1}$.

The correct answers are:

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two missiles must be equal to the total distance when the missiles meet:

This question is about distances, speeds, and times. The equation connecting these values is

We want to know the amount of time needed for the missiles to meet. Let the time taken be $t$. Then we can write an expression for the distance each of the missiles travels:

For missile 1:

For missile 2:

Now we have three different equations. We can combine them using substitution to solve for the value of $t$.

The missiles will meet after $\text{2}$ hours.

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two boats must be equal to the total distance when the boats meet:

This question is about distances, speeds, and times. The equation connecting these values is

We want to know the amount of time needed for the boats to meet. Let the time taken be $t$. Then we can write an expression for the distance each of the boats travels:

For boat 1:

For boat 2:

Now we have three different equations. We can combine them using substitution to solve for the value of $t$.

The boats will meet after $\text{7}$ hours.

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two ships must be equal to the total distance when the ships meet:

This question is about distances, speeds, and times. The equation connecting these values is

We want to know the amount of time needed for the ships to meet. Let the time taken be $t$. Then we can write an expression for the distance each of the ships travels:

For ship 1:

For ship 2:

Now we have three different equations. We can combine them using substitution to solve for the value of $t$.

The ships will meet after $\text{4}$ hours.

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

The two things we want to know in this question are the prices for each of the items (a waffle and a hotdog). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{6}$ waffles and $\text{4}$ hotdogs is $\text{N}\llap{=}\,\text{270}$." We can use the expression $6 w$ to represent the price of the $\text{6}$ waffles. Similarly, the expression $4 h$ represents the price of the $\text{4}$ hotdogs. With these values we can write a full equation for the prices:

Now we can use the second point. It says, "a waffle costs $\text{N}\llap{=}\,\text{5}$ more than a hotdog." We can write this as an equation like this:

Now that we have two equations, we need to solve them simultaneously.

We could use elimination, but substitution is a better choice (because $w$ is already the subject). Substitute the second equation into the first equation and solve!

This means that the price of one hotdog is $\text{N}\llap{=}\,\text{24}$.

Finally, use the value we found for $h$ to find the value of $w$ (the price of a waffle). We can use either equation to do this, but the second one is easier to use (because $w$ is already isolated).

The price for one waffle is $\text{N}\llap{=}\,\text{29}$.

It is important to write the answer to a word problem as a complete sentence.

The price of the waffle is $\text{N}\llap{=}\,\text{29}$ while a hotdog costs $\text{N}\llap{=}\,\text{24}$.

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

The two things we want to know in this question are the prices for each of the items (a pizza and a sandwich). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{4}$ pizzas and $\text{2}$ sandwiches is $\text{N}\llap{=}\,\text{176}$." We can use the expression $4 p$ to represent the price of the $\text{4}$ pizzas. Similarly, the expression $2 s$ represents the price of the $\text{2}$ sandwiches. With these values we can write a full equation for the prices:

Now we can use the second point. It says, "a pizza costs $\text{N}\llap{=}\,\text{8}$ more than a sandwich." We can write this as an equation like this:

Now that we have two equations, we need to solve them simultaneously.

We could use elimination, but substitution is a better choice (because $p$ is already the subject). Substitute the second equation into the first equation and solve!

This means that the price of one sandwich is $\text{N}\llap{=}\,\text{24}$.

Finally, use the value we found for $s$ to find the value of $p$ (the price of a pizza). We can use either equation to do this, but the second one is easier to use (because $p$ is already isolated).

The price for one pizza is $\text{N}\llap{=}\,\text{32}$.

It is important to write the answer to a word problem as a complete sentence.

The price of the pizza is $\text{N}\llap{=}\,\text{32}$ while a sandwich costs $\text{N}\llap{=}\,\text{24}$.

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

The two things we want to know in this question are the prices for each of the items (a sandwich and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{6}$ sandwiches and $\text{4}$ hamburgers is $\text{N}\llap{=}\,\text{330}$." We can use the expression $6 s$ to represent the price of the $\text{6}$ sandwiches. Similarly, the expression $4 h$ represents the price of the $\text{4}$ hamburgers. With these values we can write a full equation for the prices:

Now we can use the second point. It says, "a sandwich costs $\text{N}\llap{=}\,\text{5}$ more than a hamburger." We can write this as an equation like this:

Now that we have two equations, we need to solve them simultaneously.

We could use elimination, but substitution is a better choice (because $s$ is already the subject). Substitute the second equation into the first equation and solve!

This means that the price of one hamburger is $\text{N}\llap{=}\,\text{30}$.

Finally, use the value we found for $h$ to find the value of $s$ (the price of a sandwich). We can use either equation to do this, but the second one is easier to use (because $s$ is already isolated).

The price for one sandwich is $\text{N}\llap{=}\,\text{35}$.

It is important to write the answer to a word problem as a complete sentence.

The price of the sandwich is $\text{N}\llap{=}\,\text{35}$ while a hamburger costs $\text{N}\llap{=}\,\text{30}$.

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be $\text{N}\llap{=}\,\text{6.75}$. Or $\text{N}\llap{=}\,\text{5}$. Or $\text{N}\llap{=}\,\text{19.87}$. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like $\text{N}\llap{=}\,\text{4.30}$, or must a price always be a whole number like $\text{N}\llap{=}\,\text{4.00}$? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -$\text{N}\llap{=}\,\text{5.50}$. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

1. Yes, it can be a decimal
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be $\text{N}\llap{=}\,\text{6.75}$. Or $\text{N}\llap{=}\,\text{5}$. Or $\text{N}\llap{=}\,\text{19.87}$. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like $\text{N}\llap{=}\,\text{4.30}$, or must a price always be a whole number like $\text{N}\llap{=}\,\text{4.00}$? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -$\text{N}\llap{=}\,\text{5.50}$. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

1. Yes, it can be a decimal
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the number of mangoes Zenzile buys. Can the number of items be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be $\text{N}\llap{=}\,\text{6.75}$. Or $\text{N}\llap{=}\,\text{5}$. Or $\text{N}\llap{=}\,\text{19.87}$. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the number of items can be a decimal number or not. In other words, can the number of mangoes be a decimal number like $\text{3.7}$ mangoes? Or must it be a whole number like $\text{4}$ mangoes? This is different from the airtime example above: airtime can be a decimal number, but it is not possible to buy $\text{3.5}$ mangoes! (You might think, "Wait! I can buy $\text{3.5}$ mangoes, just cut one of them in half." But no one should expect to do this while shopping.)

The answer for the first question is: No, the number of items cannot be a decimal number.

The second question is similar, but this time we need to decide if the number of items can be negative, or must be postive. The number of mangoes cannot be negative. If Zenzile did not buy any mangoes at all, then the number of mangoes is zero. But is it impossible for him to buy $-\text{3}$ mangoes.

The answer to the second question is: No, the number of items cannot be a negative number.

The correct answer choices are:

1. No, it must be an integer
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Thobeka's diagram. Can the number of rectangles be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be $-\text{150}$ days of school, shame. Similarly, it is not possible to have $\text{176.3562}$ days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like $\text{3.7}$, or must the number be a whole number like $\text{4}$? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. $\text{3.7}$ rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get $\text{3.7}$ rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Thobeka drew any rectangles at all, then the answer to this question must be positive. There is no such thing as $-\text{4}$ rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

1. No, it must be an integer
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Latifah's diagram. Can the number of rectangles be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be $-\text{150}$ days of school, shame. Similarly, it is not possible to have $\text{176.3562}$ days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like $\text{3.7}$, or must the number be a whole number like $\text{4}$? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. $\text{3.7}$ rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get $\text{3.7}$ rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Latifah drew any rectangles at all, then the answer to this question must be positive. There is no such thing as $-\text{4}$ rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

1. No, it must be an integer
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be $-\text{150}$ days of school, shame. Similarly, it is not possible to have $\text{176.3562}$ days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like $\text{3.5}\ \text{cm}$, or must a distance be a whole number like $\text{4}\ \text{cm}$? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have $\text{3.5}$ days of school, but a rectangle certainly can have a width of $\text{3.5}\ \text{cm}$. Or $\text{5.927}\ \text{cm}$. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room $-\text{5}$ metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

1. Yes, it can be a decimal
2. No, it cannot be negative

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

This means that Akinlabi's mark is $\text{82}$.

Now we can use Akinlabi's mark and one of the equations we have to find Chinedu's mark. The easiest way to do that is to substitute Akinlabi's mark back into the first equation:

The students achieved these marks: Chinedu earned $\text{88}$ and Akinlabi earned $\text{82}$.

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

This means that Latifah's mark is $\text{83}$.

Now we can use Latifah's mark and one of the equations we have to find Abodunrin's mark. The easiest way to do that is to substitute Latifah's mark back into the first equation:

The students achieved these marks: Abodunrin earned $\text{91}$ and Latifah earned $\text{83}$.

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

This means that Yogo's mark is $\text{46}$.

Now we can use Yogo's mark and one of the equations we have to find Mkhuseli's mark. The easiest way to do that is to substitute Yogo's mark back into the first equation:

The students achieved these marks: Mkhuseli earned $\text{66}$ and Yogo earned $\text{46}$.

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of $\text{36}$. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the even integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the sum of the numbers is $\text{36}$". Remember that sum means addition. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of $n_{2}$.

We can solve this using substitution. However, remember that we want the value of $n_2$ (the larger number). We can start by rearranging the first equation to make $n_1$ the subject. Then the substitution step will remove $n_1$ from the second equation and we can solve for $n_2$. (This is not required - it just makes the solution faster.)

Now substitute $n_2 - 2$ into the other equation and solve for $n_2$.

The result is $n_{2} = 19$. Notice that this means that the other number, $n_{1}$, must be $\text{17}$, because $n_{2} = n_{1} + 2$. This is perfect, because we know that the sum of the numbers is $\text{36}$, and $19 + 17 = 36$.

The larger number is $\text{19}$.

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of $\text{40}$. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the even integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the sum of the numbers is $\text{40}$". Remember that sum means addition. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of $n_{1}$.

We can do this using substitution. If we substitute $n_2 = n_1 + 2$ into the equation $n_1 + n_2 = 40$, the $n_2$ terms will disappear, leaving $n_1$. Then we can solve for $n_1$, which is exactly what we want.

The result is $n_{1} = 19$. Notice that this means that the other number, $n_{2}$, must be $\text{21}$, because $n_{2} = n_{1} + 2$. This is perfect, because we know that the sum of the numbers is $\text{40}$, and $19 + 21 = 40$.

The smaller number is $\text{19}$.

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of $\text{16}$. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the even integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the sum of the numbers is $\text{16}$". Remember that sum means addition. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of $n_{1}$.

We can do this using substitution. If we substitute $n_2 = n_1 + 2$ into the equation $n_1 + n_2 = 16$, the $n_2$ terms will disappear, leaving $n_1$. Then we can solve for $n_1$, which is exactly what we want.

The result is $n_{1} = 7$. Notice that this means that the other number, $n_{2}$, must be $\text{9}$, because $n_{2} = n_{1} + 2$. This is perfect, because we know that the sum of the numbers is $\text{16}$, and $7 + 9 = 16$.

The smaller number is $\text{7}$.

You need to find Suleiman's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.

We need to determine Suleiman's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Akeju and Suleiman. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

Now let's write equations using these variables. From the question we know that: "Suleiman is $\text{11}$ years younger than his sister, Akeju." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

We also know that: "In $\text{9}$ years, Akeju will be $\text{2}$ times as old as Suleiman." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

These are the ages at which Akeju will be $\text{2}$ times as old as Suleiman. We can put all this together as follows:

Now we have to solve two equations, both of them including the variables $a_{A}$ and $a_{S}$. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is $a_S = a_A - 11$. We can substitute this into the second equation and solve for $a_{A}$.

Terrific: this means that Akeju is $\text{13}$ years old. But the question asked for us to find Suleiman's age. We can find it using the first equation, which relates the two ages:

So we finally got the answer to the question. Suleiman is $\text{2}$ years old.

The correct answer is: 2.

You need to find Richard's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.

We need to determine Richard's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Matthew and Richard. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

Now let's write equations using these variables. From the question we know that: "Matthew is $\text{7}$ years older than his brother, Richard." We need to translate that into mathematics. The key word is older, which tells us to use addition to relate the ages.

We also know that: "In $\text{6}$ years, Matthew will be $\text{8}$ times as old as Richard." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

These are the ages at which Matthew will be $\text{8}$ times as old as Richard. We can put all this together as follows:

Now we have to solve two equations, both of them including the variables $a_{M}$ and $a_{R}$. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is $a_M = a_R + 7$. We can substitute this into the second equation and solve for $a_{R}$.

Terrific: we have the answer.

But wait a minute: $a_R$ represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.

You need to find Neziswa's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.

We need to determine Neziswa's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Fezekile and Neziswa. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

Now let's write equations using these variables. From the question we know that: "Neziswa is $\text{7}$ years younger than her brother, Fezekile." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

We also know that: "In $\text{8}$ years, Fezekile will be $\text{2}$ times as old as Neziswa." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

These are the ages at which Fezekile will be $\text{2}$ times as old as Neziswa. We can put all this together as follows:

Now we have to solve two equations, both of them including the variables $a_{F}$ and $a_{N}$. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is $a_N = a_F - 7$. We can substitute this into the second equation and solve for $a_{F}$.

Terrific: this means that Fezekile is $\text{6}$ years old. But the question asked for us to find Neziswa's age. We can find it using the first equation, which relates the two ages:

So we finally got the answer to the question. But wait a minute: $a_N$ represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.

This question is about two unknown numbers. And we need to find the larger number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of $\text{35}$. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the even integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the product of the numbers is $\text{35}$". Remember that product means multiplication. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of $n_{2}$.

Remember that we want the value of $n_2$ (the larger number). We can start by rearranging the first equation to make $n_1$ the subject. Then the substitution step will remove $n_1$ from the second equation and we can solve for $n_2$. (This is not required - it just makes the solution faster.)

Now substitute $n_2 - 2$ into the other equation and solve for $n_2$.

This solution led to two answers for $n_{2}$. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves $n_{2} = 7$.

This means the other number, $n_{1}$, must be $\text{5}$, because $n_{2} = n_{1} + 2$. This is perfect, because we also know that the product of the numbers is $\text{35}$, and $7 \cdot 5 = 35$.

The larger number is $\text{7}$.

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of $\text{35}$. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the even integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the product of the numbers is $\text{35}$". Remember that product means multiplication. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of $n_{1}$.

If we subsitute $n_2 = n_1 + 2$ into the equation $n_1 \cdot n_2 = 35$, the $n_2$ terms will disappear, leaving $n_1$. Then we can solve for $n_1$, which is exactly what we want.

This solution led to two answers for $n_{1}$. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves $n_{1} = 5$.

This means the other number, $n_{2}$, must be $\text{7}$, because $n_{2} = n_{1} + 2$. This is perfect, because we also know that the product of the numbers is $\text{35}$, and $5 \cdot 7 = 35$.

The smaller number is $\text{5}$.

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of $\text{99}$. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the even integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the product of the numbers is $\text{99}$". Remember that product means multiplication. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of $n_{1}$.

If we subsitute $n_2 = n_1 + 2$ into the equation $n_1 \cdot n_2 = 99$, the $n_2$ terms will disappear, leaving $n_1$. Then we can solve for $n_1$, which is exactly what we want.

This solution led to two answers for $n_{1}$. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves $n_{1} = 9$.

This means the other number, $n_{2}$, must be $\text{11}$, because $n_{2} = n_{1} + 2$. This is perfect, because we also know that the product of the numbers is $\text{99}$, and $9 \cdot 11 = 99$.

The smaller number is $\text{9}$.

You should represent the numbers in the question by $n$ and $n+1$. Write an equation based on the fact that the product of those two numbers is $\text{156}$.

This question asks us to find two numbers which have a product of $\text{156}$. That means if we multiply the numbers we get $\text{156}$.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, $n$ is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

The question tells us that the product of these numbers is $\text{156}$. In other words, if we multiply the numbers, we get $\text{156}$. As an equation this is:

Now we can use the fact that the numbers are consecutive. So if $n_{1}$ represents the first number, then the next number must be one more than $n_{1}$.

Now we have two equations about the two numbers.

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for $n_{1}$.

The solutions to the equation are $n_{1} = -13$ or $n_{1} = 12$. This means the first number is either $\text{−13}$ or $\text{12}$.

Wait a minute! The numbers $\text{−13}$ and $\text{12}$ are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the $n_{1} = -13$, so the first number is $n_{1} = 12$.

Based on that we can find the second number. The second number is $n_{1} + 1$, which is equal to $\left(12\right) + 1 = 13$.

The two consecutive integers are $\text{12}$ and $\text{13}$.

You should represent the numbers in the question by $n$ and $n+1$. Write an equation based on the fact that the product of those two numbers is $\text{132}$.

This question asks us to find two numbers which have a product of $\text{132}$. That means if we multiply the numbers we get $\text{132}$.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, $n$ is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

The question tells us that the product of these numbers is $\text{132}$. In other words, if we multiply the numbers, we get $\text{132}$. As an equation this is:

Now we can use the fact that the numbers are consecutive. So if $n_{1}$ represents the first number, then the next number must be one more than $n_{1}$.

Now we have two equations about the two numbers.

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for $n_{1}$.

The solutions to the equation are $n_{1} = -12$ or $n_{1} = 11$. This means the first number is either $\text{−12}$ or $\text{11}$.

Wait a minute! The numbers $\text{−12}$ and $\text{11}$ are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the $n_{1} = 11$, so the first number is $n_{1} = -12$.

Based on that we can find the second number. The second number is $n_{1} + 1$, which is equal to $\left(-12\right) + 1 = -11$.

The two consecutive integers are $\text{−12}$ and $\text{−11}$.

You should represent the numbers in the question by $n$ and $n+1$. Write an equation based on the fact that the product of those two numbers is $\text{90}$.

This question asks us to find two numbers which have a product of $\text{90}$. That means if we multiply the numbers we get $\text{90}$.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, $n$ is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

The question tells us that the product of these numbers is $\text{90}$. In other words, if we multiply the numbers, we get $\text{90}$. As an equation this is:

Now we can use the fact that the numbers are consecutive. So if $n_{1}$ represents the first number, then the next number must be one more than $n_{1}$.

Now we have two equations about the two numbers.

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for $n_{1}$.

The solutions to the equation are $n_{1} = -10$ or $n_{1} = 9$. This means the first number is either $\text{−10}$ or $\text{9}$.

Wait a minute! The numbers $\text{−10}$ and $\text{9}$ are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the $n_{1} = -10$, so the first number is $n_{1} = 9$.

Based on that we can find the second number. The second number is $n_{1} + 1$, which is equal to $\left(9\right) + 1 = 10$.

The two consecutive integers are $\text{9}$ and $\text{10}$.

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.

In this question we want to find the age of two people: Bonani and his son, Abdulai. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Bonani is now $\text{5}$ times as old as Abdulai." As an equation, this is:

The second piece of information says that in "$\text{8}$ years... Bonani will be $\text{3}$ times as old as his son." In $\text{8}$ years Bonani will be $b + \text{8}$ years old, and similarly Abdulai will be $a + \text{8}$ years old. Then:

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel $b$.)

Great! Now we know that Abdulai is $\text{8}$ years old.

We can now find Bonani's age. Substitute Abdulai's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

Write your final answer: Bonani is $\text{40}$ years old and Abdulai is $\text{8}$ years old.

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.

In this question we want to find the age of two people: Abayomi and his son, Mark. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Abayomi is now $\text{8}$ times as old as Mark." As an equation, this is:

The second piece of information says that in "$\text{10}$ years... Abayomi will be $\text{3}$ times as old as his son." In $\text{10}$ years Abayomi will be $a + \text{10}$ years old, and similarly Mark will be $m + \text{10}$ years old. Then:

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel $a$.)

Great! Now we know that Mark is $\text{4}$ years old.

We can now find Abayomi's age. Substitute Mark's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

Write your final answer: Abayomi is $\text{32}$ years old and Mark is $\text{4}$ years old.

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.

In this question we want to find the age of two people: Daniel and his son, Babatunde. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Daniel is now $\text{9}$ times as old as Babatunde." As an equation, this is:

The second piece of information says that in "$\text{5}$ years... Daniel will be $\text{5}$ times as old as his son." In $\text{5}$ years Daniel will be $d + \text{5}$ years old, and similarly Babatunde will be $b + \text{5}$ years old. Then:

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel $d$.)

Great! Now we know that Babatunde is $\text{5}$ years old.

We can now find Daniel's age. Substitute Babatunde's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

Write your final answer: Daniel is $\text{45}$ years old and Babatunde is $\text{5}$ years old.

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. And non-integer values are acceptable for time values. (In other words, an amount of time does not have to be an integer.)

The correct answer is: Yes, it can be a non-integer.

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: what was the winner's average speed for the marathon. That means we need a speed value. And non-integer values are acceptable for speed values. (In other words, an amount of speed does not have to be an integer.)

The correct answer is: Yes, it can be a non-integer.

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. And non-integer values are acceptable for time values. (In other words, an amount of time does not have to be an integer.)

The correct answer is: Yes, it can be a non-integer.

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: $d$ is the diagonal, $w$ is the width, and $l$ is the length.

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is $\text{20} \ \text{cm}$ more than its width:

The length of the rectangle is $\text{10}\ \text{cm}$ more than its width:

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

And now we can substitute in the equations from above. Specifically we can substitute $w + \text{20}$ in for $d$ and $w + \text{10}$ in for $l$. Then expand the binomials and simplify the equation as much as possible.

We have a quadratic equation in standard form! It is time to solve it.

This means that the width of the rectangle is either $\text{30} \ \text{cm}$ or $-\text{10} \ \text{cm}$. But the dimensions of a rectangle cannot be negative, so $w = \text{30} \ \text{cm}$.

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects $l$ to $w$.

The width of the rectangle is $\text{30} \ \text{cm}$ and the length is $\text{40} \ \text{cm}$.

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: $d$ is the diagonal, $w$ is the width, and $l$ is the length.

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is $\text{22} \ \text{cm}$ more than its width:

The length of the rectangle is $\text{11}\ \text{cm}$ more than its width:

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

And now we can substitute in the equations from above. Specifically we can substitute $w + \text{22}$ in for $d$ and $w + \text{11}$ in for $l$. Then expand the binomials and simplify the equation as much as possible.

We have a quadratic equation in standard form! It is time to solve it.

This means that the width of the rectangle is either $\text{33} \ \text{cm}$ or $-\text{11} \ \text{cm}$. But the dimensions of a rectangle cannot be negative, so $w = \text{33} \ \text{cm}$.

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects $l$ to $w$.

The width of the rectangle is $\text{33} \ \text{cm}$ and the length is $\text{44} \ \text{cm}$.

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: $d$ is the diagonal, $w$ is the width, and $l$ is the length.

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is $\text{30} \ \text{cm}$ more than its width:

The length of the rectangle is $\text{15}\ \text{cm}$ more than its width:

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

And now we can substitute in the equations from above. Specifically we can substitute $w + \text{30}$ in for $d$ and $w + \text{15}$ in for $l$. Then expand the binomials and simplify the equation as much as possible.

We have a quadratic equation in standard form! It is time to solve it.

This means that the width of the rectangle is either $\text{45} \ \text{cm}$ or $-\text{15} \ \text{cm}$. But the dimensions of a rectangle cannot be negative, so $w = \text{45} \ \text{cm}$.

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects $l$ to $w$.

The width of the rectangle is $\text{45} \ \text{cm}$ and the length is $\text{60} \ \text{cm}$.

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable $a$ for Ayodeji's mark and $b$ for Babatunde's mark. So we can break up the first fact like this:

The first fact is equivalent to this equation: $a + b = 146$.

Similarly, we can identify key parts of the second fact, and translate each into an expression.

This equation, $a = b - 22$, means that Ayodeji's mark is less that Babatunde's mark.

The correct answers are:

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable $a$ for Ayobami's mark and $b$ for Bridget's mark. So we can break up the first fact like this:

The first fact is equivalent to this equation: $a + b = 140$.

Similarly, we can identify key parts of the second fact, and translate each into an expression.

This equation, $a = b - 12$, means that Ayobami's mark is less that Bridget's mark.

The correct answers are:

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable $a$ for Anuoluwapo's mark and $b$ for Babangida's mark. So we can break up the first fact like this:

The first fact is equivalent to this equation: $a + b = 170$.

Similarly, we can identify key parts of the second fact, and translate each into an expression.

This equation, $a = b - 10$, means that Anuoluwapo's mark is less that Babangida's mark.

The correct answers are: